unit 2 • quadratic functions and modeling lesson 5: transforming functions instruction ·...

UNIT 2 • QUADRATIC FUNCTIONS AND MODELINGLesson 5: Transforming Functions

Instruction

© Walch EducationU2-201

UCSS Secondary Mathematics II Teacher Resource 2.5.1

IntroductionYou can change a function’s position or shape by adding or multiplying a constant to that function. This is called a transformation. When adding a constant, you can transform a function in two distinct ways. The first is a transformation on the independent variable of the function; that is, given a function f(x), we add some constant k to x: f(x) becomes f(x + k). The second is a transformation on the dependent variable; given a function f(x), we add some constant k to f(x): f(x) becomes f(x) + k.

In this lesson, we consider the transformation on a function by a constant k, either when k is added to the independent variable, x, or when k is added to the dependent variable, f(x). Given f(x) and a constant k, we will observe the transformations f(x) + k and f(x + k), and examine how transformations affect the vertex of a quadratic equation.

Key Concepts

• To determine the effects of the constant on a graph, compare the vertex of the original function to the vertex of the transformed function.

• Neither f(x + k) nor f(x) + k will change the shape of the function so long as k is a constant.

• Transformations that do not change the shape or size of the function but move it horizontally and/or vertically are called translations.

• Translations are performed by adding a constant to the independent or dependent variable.

Prerequisite Skills

This lesson requires the use of the following skills:

• graphing quadratic functions

• evaluating quadratic functions


Instruction

U2-202© Walch EducationUCSS Secondary Mathematics II Teacher Resource

2.5.1

Vertical Translations—Adding a Constant to the Dependent Variable, f(x) + k

• f(x) + k moves the graph of the function k units up or down depending on whether k is greater than or less than 0.

• If k is positive in f(x) + k, the graph of the function will be moved up.

• If k is negative in f(x) + k, the graph of the function will be moved down.

Vertical translations: f(x) + k

When k is positive, k > 0, the graph moves up:

f(x) + k

f(x)

When k is negative, k < 0, the graph moves down:

f(x) + k

f(x)

Horizontal Translations—Adding a Constant to the Independent Variable, f(x + k)

• f(x + k) moves the graph of the function k units to the right or left depending on whether k is greater than or less than 0.

• If k is positive in f(x + k), the function will be moved to the left.

• If k is negative in f(x + k), the function will be moved to the right.


Instruction



Common Errors/Misconceptions

• incorrectly moving the graph in the direction opposite that indicated by k, especially in horizontal shifts; for example, moving the graph left when it should be moved right

• incorrectly moving the graph left and right versus up and down (and vice versa) when operating with f(x + k) and f(x) + k

Horizontal translations: f(x + k)

When k is positive, k > 0, the graph moves left:

f(x + k) f(x)

When k is negative, k < 0, the graph moves right:

f(x + k)f(x)


Instruction


2.5.1

Example 1

Consider the function f(x) = x2 and the constant k = 2. What is f(x) + k? How are the graphs of f(x) and f(x) + k different?

1. Substitute the value of k into the function.

If f(x) = x2 and k = 2, then f(x) + k = x2 + 2.

2. Use a table of values to graph the functions on the same coordinate plane.

x f (x) f (x) + 2–2 4 6–1 1 30 0 21 1 32 4 6

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10 f(x) + 2

f(x)

Guided Practice 2.5.1


Instruction



3. Compare the graphs of the functions.

Notice the shape and horizontal position of the two graphs are the same. The only difference between the two graphs is that the value of f(x) + 2 is 2 more than f(x) for all values of x. In other words, the transformed graph is 2 units up from the original graph.

Example 2

Consider the function f(x) = x2 and the constant k = –3. What is f(x) + k? How are the graphs of f(x) and f(x) + k different?


If f(x) = x2 and k = –3, then f(x) + k = x2 – 3.

2. Use a table of values to graph the two functions on the same coordinate plane.

x f (x) f (x) – 3–4 16 13–2 4 10 0 –32 4 14 16 13

(continued)


Instruction


2.5.1

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x) – 3

f(x)


Notice both the shape and horizontal position of the two graphs are the same. The only difference between the two graphs is that the value of f(x) – 3 is 3 less than f(x) for all values of x.

The easiest point to analyze on the graphs is the vertex. For f(x), the vertex occurs at (0, 0). For the graph of f(x) – 3, the vertex occurs at (0, –3). The transformed graph has moved down 3 units.


Instruction



Example 3

Consider the function f(x) = x2, its graph, and the constant k = 4. What is f(x + k)? How are the graphs of f(x) and f(x + k) different?


If f(x) = x2 and k = 4, then f(x + k) = f(x + 4) = (x + 4)2.

2. Use a table of values to graph the two functions on the same coordinate plane.

x f (x) f (x + 4)–6 36 4–4 16 0–2 4 40 0 162 4 364 16 64

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

123456789

10f(x + 4) f(x)


Instruction


2.5.1


Notice the shape and vertical position of the two graphs are the same. The only difference between the two graphs is that every point on the curve of f(x) has been shifted 4 units to the left in the graph of f(x + 4).

Example 4

Consider the function f(x) = x2 and the constant k = –1. What is f(x + k)? How are the graphs of f(x) and f(x + k) different?


If f(x) = x2 and k = –1, then f(x + k) = f(x – 1) = (x – 1)2.

2. Using vertex or standard form, graph the two functions on the same coordinate plane.

Recall from Example 2 that the vertex of the parent function, f(x) = x2, is (0, 0).

The vertex of the transformed function, f(x – 1) = (x – 1)2, is (1, 0). This

can be verified by using − −

b

af

b

a2 2, .

To use this formula, expand the function so that it’s in the form of a quadratic, then find a, b, and c.

f(x – 1) = (x – 1)2 = x2 – 2x + 1

a = 1, b = –2, and c = 1

The x-coordinate of the vertex is given by 2

( 2)

2(1)1=

−=− −

=xb

a.

Substitute the x-value of the vertex into the function.

f(x – 1) = (x – 1)2

f(1) = [(1) – 1]2

f(1) = 0

Therefore, the vertex is (1, 0).(continued)


Instruction



Graph the functions.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-10-9-8-7-6-5-4-3-2-1

123456789

10

f(x – 1)f(x)


Notice both the shape and vertical position of the two graphs are the same. The only difference between the two graphs is that every point on the curve of f(x) has been shifted 1 unit to the right in the graph of f(x – 1).


Instruction


2.5.1

Example 5

The revenue function for a model helicopter company is modeled by the curve f(x) = –5x2 + 400x, where x is the number of helicopters built per month and f(x) is the revenue. The owner wants to include rent in the revenue equation to determine the company’s profit per month. The company pays $2,250 per month to rent its warehouse. In terms of f(x), what equation now describes the company’s profit per month? Compare the vertices of the original function and the transformed function.

1. Build the new function.

The rent is being subtracted from the entire function. Therefore, the function follows the format f(x) + k, where k is negative.

The company makes f(x) = –5x2 + 400x dollars per month. The company also pays $2,250 dollars per month in rent. Therefore, the new function that describes the company’s profit, P(x), is as follows:

P(x) = f(x) – rent = f(x) – 2250 = –5x2 + 400x – 2250

2. Determine the vertex of f(x).

The vertex of f(x) has the x-coordinate −

=−

−=−−

=b

a2

400

2 5

400

1040

( )

( ).

Evaluate f(x) at the x-coordinate of the vertex to find the y-coordinate.

f(x) = –5x2 + 400x

f(40) = –5(40)2 + 400(40)

f(40) = –8000 + 16,000

f(40) = 8000

The vertex of f(x) is (40, 8000).


Instruction



3. Determine the vertex of P(x).

The vertex of P(x) has the same x-coordinate as f(x) since the values of a and b are the same. The x-coordinate of the vertex of P(x) is 40.

Evaluate P(x) at the x-coordinate of the vertex to find the y-coordinate.

P(x) = –5x2 + 400x – 2250

P(40) = –5(40)2 + 400(40) – 2250

P(40) = 8000 – 2250

P(40) = 5750

The vertex of P(x) is (40, 5750).

4. Compare the vertices.

The vertex of P(x) is 2,250 units lower than the vertex of f(x). The model was shifted down 2,250 units.

unit 2 • quadratic functions and modeling lesson 5: transforming functions instruction ·...

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